Solution:

Place $A = (0,0,0)$ and $E = (8,8,8),$ so that $I = (6,8,8),$ $J = (8,6,8),$ $K = (8,8,6),$ and $\overline{AE}$ has direction $(1,1,1).$ The line through $I$ in that direction leaves the cube at $L = (0,2,2);$ similarly $J$ and $K$ lead to $M = (2,0,2)$ and $N = (2,2,0).$ The tunnel wall through $I$ and $J$ is the plane $2z = x + y + 2,$ which also contains $L$ and $M$ and crosses the $z$-axis at $O = (0,0,1);$ the other two walls behave symmetrically, crossing the $y$- and $x$-axes at $(0,1,0)$ and $(1,0,0).$

Now add up the surface. Each of the three cube faces at $E$ loses a right triangle with legs $2$ (such as $IEJ$), leaving area $64 - 2 = 62.$ Each of the three faces at $A$ loses a quadrilateral of area $2:$ on the face $z = 0$ its vertices are $(0,0,0),$ $(1,0,0),$ $(2,2,0),$ $(0,1,0).$ Each tunnel wall is a pentagon like $ILOMJ:$ the rectangle $ILMJ$ with $IJ = 2\sqrt{2}$ and $IL = 6\sqrt{3}$ has area $12\sqrt{6},$ and the isosceles triangle $LOM$ with base $LM = 2\sqrt{2}$ and height $\sqrt{3}$ adds $\sqrt{6},$ for $13\sqrt{6}$ per wall.

The total surface area is $6 \cdot 62 + 3 \cdot 13\sqrt{6} = 372 + 39\sqrt{6},$ so $m + n + p = 372 + 39 + 6 = 417.$